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Full text of "Treatise On Analysis Vol-Ii"

182       XIII    INTEGRATION

(13.13.5)   Ifg is locally \i-integr able > then so is \g\9 and we have \g - /z| = \g\  JJL.

The first assertion follows immediately from (13.13.1) and (13.7.4). To
prove the second, note first that if/^ 0 is a function belonging to 3fR(X)
and if u e Jfc(X) is such that \u\ g/, then we have

(ug dn g L| - \g\ dn g (

f\g\

(13.10.3), and therefore (13.3.2.1) \g - ju| :g \g\  p.. To prove the reverse
inequality, let L denote the (compact) support of/. Let A be the set of
all x e L such that g(x) ^ 0; the set A is integrable ((13.9.9) and (13.9.2)) and
therefore (13.9.1) there exists an increasing sequence of compact subsets Kn
of A such that A - (JKW is /z-negligible. Hence, by (13.8.4), for any s > 0

there exists an integer n such that I _ f\g\ d\i<* s. The same reasoning,
using the measurability of g and (13.8.4), shows that there exists a compact
subset K'n of Kw such that g \ K^ is continuous and f tf\g\ dp < 2e. Now

JA     Kn'

consider on K^ the continuous function x\-+\g(x)\/g(x); by applying (3.18.2)
and the Tietze-Urysohn theorem (4.5.1) to the real and imaginary parts of
this function, we see that there exists a function w E Jf C(X) such that w(x) =
\9(x)\l9(x) fr all x 6 K^. The function v = w  inf(l, l/\w\) (with the conven-
tion 1/0 = +00) is then continuous on X, hence belongs to JTC(X), and is
such that v(x) = \g(x)\/g(x) in K'n and \v(x)\ g 1 throughout X. Hence
\fv\ g/and

\fvg dft =       fvg d^ +          fvg dfi;

J                       JKn                        JA Kn'

but

and

so that finally

fvg dp =      f\g\ dp,

i r                       r

fvg dfi ^

I jA-Kn'                   jA-Kn'

\

fagdfi

- 4a.

Since e > 0 was arbitrary, this completes the proof.icular, if X is compact, then jS?/OCt R(X, fi) (resp. j?,OCp C(X, ju)) is